TITLE: Prove that: $\sin(x) \cos(x) \geq x-x^3$
QUESTION [10 upvotes]: Prove that the following inequality holds for $x\ge0$ :
$$\sin(x) \cos(x) \geq x-x^3$$
This is an inequality often met during my high school classes I also used for this problem yesterday. I'm interested in a non-calculus proof if this is possible.
Proof involving calculus:
Let's consider 
$$f(x) = \sin(x) \cos(x)-x+x^3$$
then
$$f'(x) = 3 x^2-2\sin^2(x)\tag1$$
$$x\ge \sin(x)\tag2$$ 
From $(1)$ and $(2)$ we immediately notice that $f'(x)\ge0$ and taking into account that $f(0)=0$ we may conclude that the inequality holds. Thanks.

REPLY [8 votes]: A Geometric Proof
The posed inequality is equivalent to $\sin(x)\ge x-x^3/4.$
Consider the wedge of the unit circle below:
$\hspace{32mm}$
The area of the whole wedge (red and green regions) is $\frac12x$, and the area of the green triangle is $\frac12\sin(x)$. Thus, we get that $\sin(x)\le x$. Furthermore, the area of the red region is $\frac12(x-\sin(x))$.
Noting that the red region is contained in the rectangle with base $2\sin(x/2)$ and height $1-\cos(x/2)$, we get that
$$
\begin{align}
\tfrac12(x-\sin(x))
&\le2\sin(x/2)(1-\cos(x/2))\\
&=4\sin(x/2)\sin^2(x/4)\\
&\le x^3/8
\end{align}
$$
which yields
$$
x-x^3/4\le\sin(x)
$$
as desired.